Question: Captain Christopher has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Tiffany and her merciless band of thieves. The Captain has probability $\dfrac{2}{5}$ of hitting the pirate ship, if his ship hasn't already been hit. If it has been hit, he will always miss. The pirate has probability $\dfrac{3}{8}$ of hitting the Captain's ship, if her ship hasn't already been hit. If it has been hit, she will always miss as well. If the Captain shoots first, what is the probability that the Captain misses the pirate ship, but the pirate hits?
Explanation: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is the Captain missing the pirate ship and event B is the pirate hitting the Captain's ship. The Captain fires first, so his ship can't be sunk before he fires his cannons. So, the probability of the Captain missing the pirate ship is $\dfrac{3}{5}$ If the Captain missed the pirate ship, the pirate has a normal chance to fire back. So, the probability of the pirate hitting the Captain's ship given the Captain missing the pirate ship is $\dfrac{3}{8}$ The probability that the Captain misses the pirate ship, but the pirate hits is then the probability of the Captain missing the pirate ship times the probability of the pirate hitting the Captain's ship given the Captain missing the pirate ship. This is $\dfrac{3}{5} \cdot \dfrac{3}{8} = \dfrac{9}{40}$